if h(x) = x − ⌊(1/x)⌋ , x>0 then h^( −1) ( x ) = ?


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Question Number 164826 by mnjuly1970 last updated on 22/Jan/22

$i f h (x) = x - ⌊ \frac{1}{x} ⌋, x > 0$ $t h e n h^{- 1} (x) = ?$
	Answered by TheSupreme last updated on 22/Jan/22

	$⌊ \frac{1}{x} ⌋ = n f o r \frac{1}{n + 1} < x < \frac{1}{n}$ $⌊ \frac{1}{x} ⌋ = 0 f o r x > 1$ $x - ⌊ \frac{1}{x} ⌋ = x - n f o r \frac{1}{n + 1} < x < \frac{1}{n}$ $y = x - n f o r \frac{1}{n + 1} < x < \frac{1}{n}$ $x = y + n f o r \frac{1}{n + 1} < x < \frac{1}{n}$ $x = y + n f o r \frac{1}{n + 1} - n < y < \frac{1}{n} - n$ $\frac{1}{n} - n = y \to 1 - n^{2} = n y \to n = \frac{- y + \sqrt{y^{2} + 4}}{2}$ $n = ⌊ \frac{- y + \sqrt{y^{2} + 4}}{2} ⌋$ $x = y + ⌊ \frac{- y + \sqrt{y^{2} + 4}}{2} ⌋$ $f (0.2) = - 4.8$ $f^{- 1} (- 4.8) = - 4.8 + ⌊ \frac{4.8 + \sqrt{4 . 8^{2} + 4}}{2} ⌋ = 0.2$ $f (0.03) = - 32.97$ $f^{- 1} (- 32.97) = 0.03$
	Answered by mahdipoor last updated on 23/Jan/22
	$1) x>1 ⇒ y>1 ⇒0<(1/x)<1 ⇒y=x−[(1/x)]=x ⇒h^(−1) (x)=x 2) x=1 ⇒ y=0 ⇒h^(−1) (0)=1 3) 0<x<1 ⇒ y<0 y=x−[(1/x)] ⇒ {y}={x−[(1/x)]} ={x}=x ⇒ h^(−1) (x)=x 1,2,3⇒h^(−1) (x)= { ((x x>1)),((1 x=0 )),(({x} x<0)) :}$
	$1) x > 1 \Rightarrow y > 1$ $\Rightarrow 0 < \frac{1}{x} < 1 \Rightarrow y = x - [\frac{1}{x}] = x \Rightarrow h^{- 1} (x) = x$ $2) x = 1 \Rightarrow y = 0 \Rightarrow h^{- 1} (0) = 1$ $3) 0 < x < 1 \Rightarrow y < 0$ $y = x - [\frac{1}{x}] \Rightarrow {y} = {x - [\frac{1}{x}]} = {x} = x \Rightarrow$ $h^{- 1} (x) = x$ $1, 2, 3 \Rightarrow h^{- 1} (x) = {\begin{cases} x x > 1 \\ 1 x = 0 \\ {x} x < 0 \end{cases}$
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